volumes by slicing: disks and washers objective: to find the volume of figures using the method of...
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Volumes by Slicing: Disks and Washers
Objective: To find the volume of figures using the method of
disks/washers.
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Volume
• Recall that the underlying principle for finding the area of a plane region is to divide the region into thin strips, approximate the area of each strip by the area of a rectangle, add the approximations to form a Riemann Sum, and take the limit of the Riemann Sums to produce an integral for the area.
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Volume
• Under appropriate conditions, the same strategy can be used to find the volume of a solid. The idea is to divide the solid into thin slabs, approximate the volume of each slab, add the approximations to form a Riemann Sum, and take the limit of the Riemann Sums to produce an integral for the volume.
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The Volume Problem
• 7.2.1 Let S be a solid that extends along the x-axis and is bounded on the left and right, respectively, by the planes that are perpendicular to the x-axis at x = a and x = b. Find the volume V of the solid, assuming that its cross-sectional area A(x) is known at each x in the interval [a, b].
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The Volume Problem
• To solve this problem, we begin by dividing the interval [a, b] into n subintervals, thereby dividing the solid into n slabs. If we assume that the width of the kth subinterval is , then the volume of the kth slab can be approximated by the volume of a right cylinder of width (height) and cross sectional area , where is a point in the kth subinterval.
xkk xxA )( *
kx)( *kxA
*kx
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The Volume Problem
• Adding these approximations yields the following Riemann Sum that approximates the volume V:
• Taking the limit as n increases and the widths of the subintervals approach zero yields the definite integral
n
kkk xxAV
1
* )(
b
a
n
kkk
xdxxAxxAV )()(lim
1
*
0max
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The Volume Formula
• 7.2.2 Let S be a solid bounded by two parallel planes perpendicular to the x-axis at x = a and x = b. If, for each x in [a, b], the cross-sectional area of S perpendicular to the x-axis is A(x), then the volume of the solid is
provided A(x) is integrable.
b
a
dxxAV )(
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The Volume Formula
• 7.2.2 Let S be a solid bounded by two parallel planes perpendicular to the y-axis at y = c and y = d. If, for each y in [c, d], the cross-sectional area of S perpendicular to the y-axis is A(y), then the volume of the solid is
provided A(y) is integrable.
d
c
dyyAV )(
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Solids of Revolution
• A solid of revolution is a solid that is generated by revolving a plane region about a line that lies in the same plane as the region; the line is called the axis of revolution. Many familiar solids are of this type.
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Volume Problem
• 7.2.4 Let f be continuous and nonnegative on [a, b], and let R be the region that is bounded above by y = f(x), below by the x-axis, and on the sides by the lines x = a and x = b. Find the volume of the solid of revolution that is generated by revolving the region R about the x-axis.
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Volume Problem
• We can solve this problem by slicing. For this purpose, observe that the cross section of the solid taken perpendicular to the x-axis at the point x is a circular disk of radius f(x).
• The area of this region is
• The volume of the solid is
2)]([)( xfxA
dxxfVb
a 2)]([
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Method of Disks
• Because the cross sections are disk shaped, the application of this formula is called the method of disks.
• The area of this region is
• The volume of the solid is
2)]([)( xfxA
dxxfVb
a 2)]([
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Example 2
• Find the volume of the solid that is obtained when the region under the curve over the interval [1, 4] is revolved about the x-axis.
xy
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Example 2
• Find the volume of the solid that is obtained when the region under the curve over the interval [1, 4] is revolved about the x-axis.
xy
2
15
2)]([
4
1
4
1
22
xxdxdxxfV
b
a
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Example 3
• Derive the formula for the volume of a sphere of radius r.
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Example 3
• Derive the formula for the volume of a sphere of radius r.
33
2222
3
4
3)()]([ r
xxrdxxrdxxfV
r
r
r
r
b
a
22 xry
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Problem
• 7.2.5 Let f and g be continuous and nonnegative on [a, b], and suppose that f(x) > g(x) for all x in the interval [a, b]. Let R be the region that is bounded above by y = f(x), below by y = g(x), and on the sides by the lines x = a and x = b. Find the volume of the solid of revolution that is generated by revolving the region R about the x-axis.
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Volumes By Washers
• We can solve this problem by slicing. For this purpose, observe that the cross section of the solid taken perpendicular to the x-axis at the point x is the annular or “washer-shaped” region with inner radius g(x) and outer radius f(x). Thus,
• The area is
• The volume is
b
a
dxxgxfV ))]([)](([ 22
22 )]([)]([)( xgxfxA
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Example 4
• Find the volume of the solid generated when the region between the graphs of the equation and over the interval [0, 2] is revolved about the x-axis.
2
2
1)( xxf
xxg )(
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Example 4
• Find the volume of the solid generated when the region between the graphs of the equation and over the interval [0, 2] is revolved about the x-axis.
2
2
1)( xxf
xxg )(
2
0
222
10
69)]5([.
dxxxV
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Perpendicular to the y-axis
• The methods or disks and washers have analogs for regions that are revolved about the y-axis. Using the method of slicing, we should have no problem deducing the following formulas for the volumes of the solids.
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Disks/Washers
• These formulas are:
d
c
dyyvywV ))]([)](([ 22d
c
dyyuV 2)]([
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Example 5
• Find the volume of the solid generated when the region enclosed by , y = 2, and x = 0 is revolved about the y-axis.
xy
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Example 5
• Find the volume of the solid generated when the region enclosed by , y = 2, and x = 0 is revolved about the y-axis.
5
32
5)(
2
0
2
0
522
ydyyV
xy
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Example 6
• Find the volume of the solid whose base is the region bounded between the curve and the x-axis from x = 0 to x = 3 whose cross sections taken perpendicular to the x-axis are:
a) squaresb) semi-circles
xy
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Example 6
• Find the volume of the solid whose base is the region bounded between the curve and the x-axis from x = 0 to x = 3 whose cross sections taken perpendicular to the x-axis are:
a) Squares• Remember, we will use the method of slicing to find
the area of a cross section and then integrate. The side of each square is the value of the function, so the area is the function squared.
2
27
2)(
3
0
23
0
2
x
dxxV
xy
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Example 6
• Find the volume of the solid whose base is the region bounded between the curve and the x-axis from x = 0 to x = 3 whose cross sections taken perpendicular to the x-axis are:
b) semi-circles • We need to find the area of a semi-circle to use the
method of slicing. The diameter of the circle is the value of the function, so the radius is half of that.
16
9
1622
3
0
223
0
xdx
xV
xy
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Example 7
• Now, we will try to rotate a figure around a line other than the x or y-axis. We will use the idea of the outer and inner radius to find the correct formula.
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Example 7
• Find the volume of the figure generated by rotating the graphs of y = 2x and y = x2 from y = 0 to y = 4 around the line x = 2.
2y x2y x
r
R
2
yx
y x
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Example 7
• Find the volume of the figure generated by rotating the graphs of y = 2x and y = x2 from y = 0 to y = 4 around the line x = 2.
• The outer radius is
• The inner radius is
• The volume is
2y x2y x
r
R
2
yx
y x
22
yR
2r y
4 2 2
0V R r dy
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Example 7
• Find the volume of the figure generated by rotating the graphs of y = 2x and y = x2 from y = 0 to y = 4 around the line x = 2.
• The outer radius is
• The inner radius is
• The volume is
2y x2y x
r
R
2
yx
y x
22
yR
2r y
2
24
02 22
yy dy
8
3